Problem: Which of the following numbers is a multiple of 3? ${41,59,72,91,107}$
Solution: The multiples of $3$ are $3$ $6$ $9$ $12$ ..... In general, any number that leaves no remainder when divided by $3$ is considered a multiple of $3$ We can start by dividing each of our answer choices by $3$ $41 \div 3 = 13\text{ R }2$ $59 \div 3 = 19\text{ R }2$ $72 \div 3 = 24$ $91 \div 3 = 30\text{ R }1$ $107 \div 3 = 35\text{ R }2$ The only answer choice that leaves no remainder after the division is $72$ $ 24$ $3$ $72$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $3$ are contained within the prime factors of $72$ $72 = 2\times2\times2\times3\times3 3 = 3$ Therefore the only multiple of $3$ out of our choices is $72$. We can say that $72$ is divisible by $3$.